We are tasked with simplifying the expression: \[ \frac{\sin 7x + \sin 5x}{\cos 7x + \cos 5x}. \] Step 1: Use the sum-to-product identities for sine and cosine.
The sum-to-product identity for sine is: \[ \sin A + \sin B = 2 \sin\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right), \] and for cosine: \[ \cos A + \cos B = 2 \cos\left(\frac{A + B}{2}\right) \cos\left(\frac{A - B}{2}\right). \] Step 2: Apply the sum-to-product identities to the given expression:
- For \( \sin 7x + \sin 5x \), we have: \[ \sin 7x + \sin 5x = 2 \sin\left(\frac{7x + 5x}{2}\right) \cos\left(\frac{7x - 5x}{2}\right) = 2 \sin(6x) \cos(x). \] - For \( \cos 7x + \cos 5x \), we have: \[ \cos 7x + \cos 5x = 2 \cos\left(\frac{7x + 5x}{2}\right) \cos\left(\frac{7x - 5x}{2}\right) = 2 \cos(6x) \cos(x). \] Step 3: Substitute these into the original expression: \[ \frac{\sin 7x + \sin 5x}{\cos 7x + \cos 5x} = \frac{2 \sin(6x) \cos(x)}{2 \cos(6x) \cos(x)}. \] Simplify by canceling out \( 2 \cos(x) \): \[ \frac{\sin(6x)}{\cos(6x)} = \tan(6x). \]
Thus, the simplified form of the expression is \( \tan 6x \).
Therefore, the correct answer is option (E).
For the reaction:
\[ 2A + B \rightarrow 2C + D \]
The following kinetic data were obtained for three different experiments performed at the same temperature:
\[ \begin{array}{|c|c|c|c|} \hline \text{Experiment} & [A]_0 \, (\text{M}) & [B]_0 \, (\text{M}) & \text{Initial rate} \, (\text{M/s}) \\ \hline I & 0.10 & 0.10 & 0.10 \\ II & 0.20 & 0.10 & 0.40 \\ III & 0.20 & 0.20 & 0.40 \\ \hline \end{array} \]
The total order and order in [B] for the reaction are respectively: